In figure , if the measure of angle $BQP$ measures $37$ degrees, determine the value of $x+y$.(Answer:$14-4\sqrt6$)
I try:
$\triangle PBQ:(3k-4k-5k) \implies 3k+4k = 2r+5k \implies 2k = 2.2 \therefore k = 2$
$\therefore PB = 6: BQ = 8: PQ =10$
$ \triangle APM: AP+2x =AM+MP(I)\\ \triangle QCN: QC+2y = NC+NQ(II)\\ \triangle ABC: AP+6+8+QC = AC+2r \implies 14 = AC+2r-AP-QC(III)\\ AC =AM+MN+NC \implies From(III):14 = AM+MN+NC+2r-AP-QC \implies AM+NC-AP-QC = 14-2r-MN(IV) \\ (I)+(II): x+y = \dfrac{(AM+MP+NC+NQ -AP-QC)}{2}(V) \\ From(IV)and(V):x+y=\dfrac{14-2r-MN+MP+NQ}{2}$
But I couldn't finish
The tricky-to-use information in the question is that $AQ, CP$ are angle bisectors.
Since $CP$ is angle bisector, $\triangle CBP \cong \triangle CMP \Rightarrow PB=PM=6$. Similarly $\triangle ABQ \cong \triangle ANQ \Rightarrow QB=QN=8$.
Let $AQ, CP$ meet at $I$, then $I$ is incenter of $\triangle ABC$. Now $PC$ is diameter hence, a line of symmetry of the big circle. As the incircle is tangent to $PB, BC$ above the diameter $PC$, it is symmetrically tangent to $PM, MC$ below diameter $PC$. Similarly conclude the circle is tangent to $BQ, QN$ symmetrically. A more accurate diagram looks like the following :
Incircle being tangent to both $PM, QN$ has diameter $2r=MN$. Length of $MN$ can be found as usual by dropping perpendicular from $P$ to $R$ on $QN$ and using Pythagoras in $\triangle PQR$. Easy enough $MN=4\sqrt{6} \Rightarrow x+y=14-4\sqrt{6}$. $\blacksquare$